Chapter 7 Isogonal Conjugates

Chapter 7

Isogonal conjugates

7.1 Directed angles

A reference triangle ABC in a plane induces an orientation of the plane, with respect to which all angles are signed. For two given lines L and L′, the directed angle ∠(L, L′) between them is the angle of rotation from L to L′ in the induced orientation of the plane. It takes values of modulo π. The following basic properties of directed angles make many geometric reasoning simple without the reference of a diagram.

Theorem 7.1. (1) ∠(L′, L)= ∠(L, L′). − (2) ∠(L1, L2)+ ∠(L2, L3)= ∠(L1, L3) for any three lines L1, L2 and L3. (3) Four points P , Q, X, Y are concyclic if and only if ∠(PX,XQ)= ∠(PY,YQ).

Remark. In calculations with directed angles, we shall slightly abuse notations by using the equality sign instead of the sign for congruence modulo π. It is understood that directed angles are deﬁned up to multiples of π. For example, we shall write β + γ = α even though it should be more properly β + γ = π α or β + γ α mod π. − − ≡− Exercise 1. If a, b, c are the sidelines of triangle ABC, then ∠(a, b)= γ etc. − 222 Isogonal conjugates

7.2 Isogonal conjugates

Let P be a given point. Consider the reﬂections of the cevians AP , BP , CP in the respec- tive bisectors of angles A, B, C, i.e., By Ceva’s theorem, these reﬂections are concurrent. Their intersection is the isogonal conjugate of P . Let P and Q be isogonal conjugates, AP and AQ intersecting BC at X and X′ respectively. Then 2 BX BX′ c = 2 . XC · X′C b Example 7.1. The incenter is the isogonal conjugate of itself. The same is true for the excenters.

Example 7.2. (The circumcenter and orthocenter) For a given triangle with circumcenter O, the line OA and the altitude through A are isogonal lines, similarly for the circumradii and altitudes through B and C. Since the circumradii are concurrent at O, the altitudes also are concurrent. Their intersection is the orthocenter H, which is the isogonal conjugate of O.

Y F E O

Z H

B X D C 7.3 The symmedian point and the centroid 223

7.3 The symmedian point and the centroid

The isogonal lines of the medians are called the symmedians. The isogonal conjugate of the centroid G is called the symmedian point K of the triangle.

G K

B C

Consider triangle ABC together with its tangential triangle A′B′C′, the triangle bounded by the tangents of the circumcircle at the vertices.

C′

B′

B D C

A′ Z

Since A′ is equidistant from B and C, we construct the circle A′(B) = A′(C) and extend the sides AB and AC to meet this circle again at Z and Y respectively. Note that

∠(A′Y,A′B′)= π 2(π α γ)= π 2β, − − − − 224 Isogonal conjugates

and similarly, ∠(A′C′,A′Z′)= π 2γ. Since ∠(A′B′,A′C′)= π 2α, we have − −

∠(A′Y,A′Z)=∠(A′Y,A′B′)+ ∠(A′B′,A′C′)+ ∠(A′C′,A′Z) =(π 2β)+(π 2α)+(π 2γ) − − − =π 0 mod π. ≡

This shows that Y , A′ and Z′ are collinear, so that (i) AA′ is a median of triangle AY Z, (ii) AY Z and ABC are similar. It follows that AA′ is the isogonal line of the A-median, i.e.,a symmedian. Similarly, the BB′ and CC′ are the symmedians isogonal to B- and C-medians. The lines AA′, BB′, CC′ therefore intersect at the isogonal conjugate of the centroid G. 7.4 Isogonal conjugates of the Gergonne and Nagel points 225

7.4 Isogonal conjugates of the Gergonne and Nagel points

7.4.1 The Gergonne point and the insimilicenter T+

Consider the intouch triangle DEF of triangle ABC.

(1) If D′ is the reﬂection of D in the bisector AI, then (i) D′ is a point on the incircle, and (ii) the lines AD and AD′ are isogonal with respect to A.

A A

E E F ′ E′ F F I Ge I P

D′ D′ B D C B D C

(2) Likewise, E′ and F ′ are the reﬂections of E and F in the bisectors BI and CI respectively, then (i) these are points on the incircle, (ii) the lines BE′ and CF ′ are isogonals of BE and CF with respect to angles B and C.

Therefore, the lines AD′, BE′, and CF ′ concur at the isogonal conjugate of the Ger- gonne point.

(3) In fact, E′F ′ is parallel to BC.

E F ′ E′

F I

D′ B D C

This follows from 226 Isogonal conjugates

(ID, IE′) =(ID, IE)+(IE, IE′) =(ID, IE)+2(IE, IB) =(ID, IE)+2((IE, AC)+(AC,IB)) π =(ID, IE)+2(AC, IB) since(IE,AC)= 2 β =(π γ) + 2 γ + − 2 =β + γ = α (mod π); − (ID, IF ′) =(ID, IF )+(IF, IF ′) =(ID, IF )+2(IF, IC) =(ID, IF )+2((IF, AB)+(AB,IC)) π =(ID, IF )+2(AB, IC) since(IF,AB)= 2 γ = (π β) 2 β + − − − 2 = (β + γ)= α (modπ) −

Since E′ and F ′ are on the incircle, and ID BC, it follows that E′F ′ is parallel to BC. ⊥ (4) Similarly, F ′D′ and D′E′ are parallel to CA and AB respectively. It follows that D′E′F ′ is homothetic to ABC. The ratio of homothety is r : R. Therefore, the center of homothety is the point T+ which divides OI in the ratio R : r. This is the internal center of similitude, or simply the insimilicenter of (O) and (I).

E F ′ E′

T+ F I O Ge

D′ B D C 7.4 Isogonal conjugates of the Gergonne and Nagel points 227

7.4.2 The Nagel point and the exsimilicenter T − The isogonal conjugate of the Nagel point is the point T which divides OI in the ratio − OT : T I = R : r. This is the external center of similitude (or exsimilicenter) of the circumcircle− − and the− incircle.

Ic D1′ Cc

O B I b T N − a

E1′ F1′

B Aa C

Ia 228 Isogonal conjugates

7.5 The Brocard points

Analogous to the Crelles points, we may ask if there are concurrent lines through the vertices making equal angles with the sidelines. More precisely, given triangle ABC, does there exist a point P satisfying ∠BAP = ∠CBP = ∠ACP = ω. It turns out that is one such unique conﬁguration.

P ω ω B C Note that if P is a point satisfying ∠BAP = ∠CBP , then the circle through P , A, B is tangent to BC at B. This circle is unique and can be constructed as follows. Its center is the intersection of the perpendicular bisector of AB and the perpendicular to BC at B.

ω B C

Likewise, if ∠CBP = ∠ACP , then the circle through P , B, C is tangent to CA at C. It follows that P is the intersection of these two circles. With this P , the circle PCA is tangent to AB at A. By Ceva’s theorem, the angle ω satisﬁes the equation sin3 ω = sin(β ω)sin(α ω)sin(γ ω). − − − It also follows that with the same ω, there is another triad of circles intersecting at another point Q such that ∠CAQ = ∠ABQ = ∠BCQ = ω. 7.5 The Brocard points 229

A A

ω ω

Q ω

P ω ω ω B C B C

The points P and Q are isogonal conjugates. They are called the Brocard points of triangle ABC.

ω ω

ω P ω ω ω B C 230 Isogonal conjugates

7.6 Kariya’s theorem

Given a triangle ABC with incenter I, consider a point X on the perpendicular from I to BC, such that IX = t. We regard t> 0 if X and the point of tangency of the incircle with the side BC are on the same side of I.

Theorem 7.2 (Kariya). Let I be the incenter of triangle ABC. If points X, Y , Z are chosen on the perpendiculars from I to BC, CA, AB respectively such that IX = IY = IZ, then the lines AX, BY , CZ are concurrent.

I Q

B C

Proof. (1) We compute the length of AX. Let the perpendicular from A to BC and the parallel from X to the same line intersect at X′. In the right triangle AXX′,

2∆ 2rs r(b + c) AX′ = r + t = r + t = + t, a − a − a XX′ = (s b) c cos B − − 1 1 = (c + a b) (c2 + a2 b2) 2 − − 2a − a(c + a b) (c2 + a2 b2) = − − − 2a b2 c2 a(b c) (b c)(b + c a) (b c)(s a) = − − − = − − = − − . 2a 2a a

Applying the Pythagorean theorem to the right triangle AXX′, we have 7.6 Kariya’s theorem 231

M ′ A A

O I I P

B C B C

X′ X X

r(b + c) 2 (b c)(s a) 2 AX2 = + t + − − a a r2(b + c)2 +(b c)2(s a)2 2r(b + c)t = − − + + t2 a2 a 2 (s a)(s b)(s c)(b+c) 2 2 − − − +(s a) (b c) 2r(b + c)t + at2 = s − − + a2 a (s a)((s b)(s c)(b + c)2 + s(s a)(b c)2) 2r(b + c)t + at2 = − − − − − + a2s a (s a) a2bc 2r(b + c)t + at2 = − · + a2s a abc(s a) 2r(b + c)t + at2 = − + as a 4Rr(s a) 2r(b + c)t + at2 = − + a a 4Rr(s a)+2r(b + c)t + at2 = − . a

(2) Let M ′ be the midpoint of the arc BAC of the circumcircle, and let MX intersect OI at P . We shall prove that angle IAP = angle IAX. First of all,

(s a)2 bc bc(s a) 4Rr(s a) AI2 = − =(s a)2 = − = − . cos2 A − · s(s a) s a 2 − For later use, we also establish

4Rr(s a) 2Rr(2s 2a)+2Rr a 2Rr(b + c) AI2 +2Rr = − +2Rr = − · = . a a a

Since IP : PO = t : R, IP = t OI. Recall that OI2 = R2 2Rr. Applying the law R+t · − 232 Isogonal conjugates of cosines to triangle AIP , we have AP 2 = AI2 + IP 2 2 AI IP cos AIP − · · t2 t = AI2 + (R2 2Rr) (AI2 + OI2 R2) (R + t)2 − − R + t − t2 t = AI2 + (R2 2Rr) (AI2 2Rr) (R + t)2 − − R + t − R 2R2rt R2t2 = AI2 + + R + t · (R + t)2 (R + t)2 R(R + t)AI2 +2R2rt + R2t2 = (R + t)2 R2 AI2 + R(AI2 +2Rr)t + R2t2 = · (R + t)2 2 2 4Rr(s a) 2R r(b+c) 2 2 R − + t + R t = · a a (R + t)2 R2(4Rr(s a)+2r(b + c)t + at2) = − . a(R + t)2

2 Note that AP 2 = R AX2. This means that AP = R AX. (R+t)2 · R+t · (3) Let AI intersect PX at X′′ and the circumcircle again at M, the antipode of M ′. Note that MM ′ = 2 and OI = R+t . M ′O IP t − − M ′

O I P

X′′

B C

X M Applying Menelaus’ theorem to triangle M ′OP and transversal IX′′M, we have

PX′′ M ′M OI PX′′ t PX′′ t 1= = = − = = − . − X′′M ′ · MO · IP ⇒ X′′M ′ 2(R + t) ⇒ PM ′ 2R + t XP t XP 2R+t Now PM = R . Therefore, PX = R , and ′ ′′ − XX R + t AX ′′ = = . X′′P R AP 7.7 Isogonal conjugate of an inﬁnite point 233

This shows that AX′′ bisects angle XAP . Since AX′′ is the bisector of angle A, the lines AP and AX are isogonal with respect to angle A. (4) Likewise, if points Y and Z are chosen on the perpendiculars from I to CA and AB such that IY = IZ = t = IX, then with the same point P on OI, the lines BP and BY are isogonal with respect to angle B, and CP , CZ isogonal with respect to angle C. Therefore the three lines AX, BY , CZ intersect at the isogonal conjugate of P (which divides OI in the ratio OP : PI = R : t).

7.7 Isogonal conjugate of an inﬁnite point

Proposition 7.3. Given a triangle ABC and a line ℓ, let ℓa, ℓb, ℓc be the parallels to ℓ through A, B, C respectively, and ℓa′ , ℓb′ , ℓc′ their reﬂections in the angle bisectors AI, BI, CI respectively. The lines ℓa′ , ℓb′ , ℓc′ intersect at a point on the circumcircle of triangle ABC.

ℓ O P b′

ℓa′ B C

ℓ ℓa ℓb ℓc ℓc′

Proof. Let P be the intersection of ℓb′ and ℓc′ .

(BP, PC) =(ℓb′ , ℓc′ )

=(ℓb′ , IB)+(IB, IC)+(IC, ℓc′ )

=(IB, ℓb)+(IB, IC)+(ℓc, IC) =(IB, ℓ)+(IB, IC)+(ℓ, IC) =2(IB, IC) π A =2 + 2 2 =(BA, AC) (mod π).

Therefore, ℓb′ and ℓc′ intersect at a point on the circumcircle of triangle ABC. Similarly, ℓa′ and ℓb′ intersect at a point P ′ on the circumcircle. Clearly, P and P ′ are the same point since they are both on the reﬂection of ℓb in the bisector IB. Therefore, the three reﬂections ℓa′ , ℓb′ , and ℓc′ intersect at the same point on the circumcircle. 234 Isogonal conjugates

Proposition 7.4. The isogonal conjugates of the inﬁnite points of two perpendicular lines are antipodal points on the circumcircle.

ℓ′ Q

ℓ I O P

B C

Proof. If P and Q are the isogonal conjugates of the inﬁnite points of two perpendicular lines ℓ and ℓ′ through A, then AP and AQ are the reﬂections of ℓ and ℓ′ in the bisector AI.

π (AP, AQ)=(AP, IA)+(IA, AQ)= (ℓ, IA) (IA, ℓ′)= (ℓ, ℓ′)= . − − − 2 Therefore, P and Q are antipodal points.